Check if a Binary Tree contains duplicate subtrees of size 2 or more

• Difficulty Level : Hard
• Last Updated : 28 Jun, 2021

Given a Binary Tree, check whether the Binary tree contains a duplicate sub-tree of size 2 or more.
Note : Two same leaf nodes are not considered as subtree size of a leaf node is one.

Input :  Binary Tree
A
/    \
B        C
/   \       \
D     E       B
/  \
D    E
Output : Yes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution. Tree with duplicate Sub-Tree [ highlight by blue color ellipse ]

[ Method 1]
A simple solution is that, we pick every node of tree and try to find is any sub-tree of given tree is present in tree which is identical with that sub-tree. Here we can use below post to find if a subtree is present anywhere else in tree.
Check if a binary tree is subtree of another binary tree

[Method 2 ]( Efficient solution )
An Efficient solution based on tree serialization and hashing. The idea is to serialize subtrees as strings and store the strings in hash table. Once we find a serialized tree (which is not a leaf) already existing in hash-table, we return true.

Below The implementation of above idea.

C++

// C++ program to find if there is a duplicate
// sub-tree of size 2 or more.
#include<bits/stdc++.h>
using namespace std;

// Separator node
const char MARKER = '\$';

// Structure for a binary tree node
struct Node
{
char key;
Node *left, *right;
};

// A utility function to create a new node
Node* newNode(char key)
{
Node* node = new Node;
node->key = key;
node->left = node->right = NULL;
return node;
}

unordered_set<string> subtrees;

// This function returns empty string if tree
// contains a duplicate subtree of size 2 or more.
string dupSubUtil(Node *root)
{
string s = "";

// If current node is NULL, return marker
if (root == NULL)
return s + MARKER;

// If left subtree has a duplicate subtree.
string lStr = dupSubUtil(root->left);
if (lStr.compare(s) == 0)
return s;

// Do same for right subtree
string rStr = dupSubUtil(root->right);
if (rStr.compare(s) == 0)
return s;

// Serialize current subtree
s = s + root->key + lStr + rStr;

// If current subtree already exists in hash
// table. [Note that size of a serialized tree
// with single node is 3 as it has two marker
// nodes.
if (s.length() > 3 &&
subtrees.find(s) != subtrees.end())
return "";

subtrees.insert(s);

return s;
}

// Driver program to test above functions
int main()
{
Node *root = newNode('A');
root->left = newNode('B');
root->right = newNode('C');
root->left->left = newNode('D');
root->left->right = newNode('E');
root->right->right = newNode('B');
root->right->right->right = newNode('E');
root->right->right->left= newNode('D');

string str = dupSubUtil(root);

(str.compare("") == 0) ? cout << " Yes ":
cout << " No " ;
return 0;
}

Java

// Java program to find if there is a duplicate
// sub-tree of size 2 or more.
import java.util.HashSet;
public class Main {

static char MARKER = '\$';

// This function returns empty string if tree
// contains a duplicate subtree of size 2 or more.
public static String dupSubUtil(Node root, HashSet<String> subtrees)
{
String s = "";

// If current node is NULL, return marker
if (root == null)
return s + MARKER;

// If left subtree has a duplicate subtree.
String lStr = dupSubUtil(root.left,subtrees);
if (lStr.equals(s))
return s;

// Do same for right subtree
String rStr = dupSubUtil(root.right,subtrees);
if (rStr.equals(s))
return s;

// Serialize current subtree
s = s + root.data + lStr + rStr;

// If current subtree already exists in hash
// table. [Note that size of a serialized tree
// with single node is 3 as it has two marker
// nodes.
if (s.length() > 3 && subtrees.contains(s))
return "";

return s;
}

//Function to find if the Binary Tree contains duplicate
//subtrees of size 2 or more
public static String dupSub(Node root)
{
HashSet<String> subtrees=new HashSet<>();
return dupSubUtil(root,subtrees);
}

public static void main(String args[])
{
Node root = new Node('A');
root.left = new Node('B');
root.right = new Node('C');
root.left.left = new Node('D');
root.left.right = new Node('E');
root.right.right = new Node('B');
root.right.right.right = new Node('E');
root.right.right.left= new Node('D');
String str = dupSub(root);
if(str.equals(""))
System.out.print(" Yes ");
else
System.out.print(" No ");
}
}

// A binary tree Node has data,
// pointer to left child
// and a pointer to right child
class Node {
int data;
Node left,right;
Node(int data)
{
this.data=data;
}
};
//This code is contributed by Gaurav Tiwari

Python3

# Python3 program to find if there is
# a duplicate sub-tree of size 2 or more

# Separator node
MARKER = '\$'

# Structure for a binary tree node
class Node:

def __init__(self, x):

self.key = x
self.left = None
self.right = None

subtrees = {}

# This function returns empty if tree
# contains a duplicate subtree of size
# 2 or more.
def dupSubUtil(root):

global subtrees

s = ""

# If current node is None, return marker
if (root == None):
return s + MARKER

# If left subtree has a duplicate subtree.
lStr = dupSubUtil(root.left)

if (s in lStr):
return s

# Do same for right subtree
rStr = dupSubUtil(root.right)

if (s in rStr):
return s

# Serialize current subtree
s = s + root.key + lStr + rStr

# If current subtree already exists in hash
# table. [Note that size of a serialized tree
# with single node is 3 as it has two marker
# nodes.
if (len(s) > 3 and s in subtrees):
return ""

subtrees[s] = 1

return s

# Driver code
if __name__ == '__main__':

root = Node('A')
root.left = Node('B')
root.right = Node('C')
root.left.left = Node('D')
root.left.right = Node('E')
root.right.right = Node('B')
root.right.right.right = Node('E')
root.right.right.left= Node('D')

str = dupSubUtil(root)

if "" in str:
print(" Yes ")
else:
print(" No ")

# This code is contributed by mohit kumar 29

C#

// C# program to find if there is a duplicate
// sub-tree of size 2 or more.
using System;
using System.Collections.Generic;

class GFG
{

static char MARKER = '\$';

// This function returns empty string if tree
// contains a duplicate subtree of size 2 or more.
public static String dupSubUtil(Node root,
HashSet<String> subtrees)
{
String s = "";

// If current node is NULL, return marker
if (root == null)
return s + MARKER;

// If left subtree has a duplicate subtree.
String lStr = dupSubUtil(root.left,subtrees);
if (lStr.Equals(s))
return s;

// Do same for right subtree
String rStr = dupSubUtil(root.right,subtrees);
if (rStr.Equals(s))
return s;

// Serialize current subtree
s = s + root.data + lStr + rStr;

// If current subtree already exists in hash
// table. [Note that size of a serialized tree
// with single node is 3 as it has two marker
// nodes.
if (s.Length > 3 && subtrees.Contains(s))
return "";

return s;
}

// Function to find if the Binary Tree contains
// duplicate subtrees of size 2 or more
public static String dupSub(Node root)
{
HashSet<String> subtrees = new HashSet<String>();
return dupSubUtil(root,subtrees);
}

// Driver code
public static void Main(String []args)
{
Node root = new Node('A');
root.left = new Node('B');
root.right = new Node('C');
root.left.left = new Node('D');
root.left.right = new Node('E');
root.right.right = new Node('B');
root.right.right.right = new Node('E');
root.right.right.left= new Node('D');
String str = dupSub(root);
if(str.Equals(""))
Console.Write(" Yes ");
else
Console.Write(" No ");
}
}

// A binary tree Node has data,
// pointer to left child
// and a pointer to right child
public class Node
{
public int data;
public Node left,right;
public Node(int data)
{
this.data = data;
}
};

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript program to find if there is a duplicate
// sub-tree of size 2 or more.

let MARKER = '\$';

// A binary tree Node has data,
// pointer to left child
// and a pointer to right child
class Node {
constructor(data)
{
this.data=data;
}
}

// This function returns empty string if tree
// contains a duplicate subtree of size 2 or more.
function dupSubUtil(root,subtrees)
{
let s = "";

// If current node is NULL, return marker
if (root == null)
return s + MARKER;

// If left subtree has a duplicate subtree.
let lStr = dupSubUtil(root.left,subtrees);
if (lStr==(s))
return s;

// Do same for right subtree
let rStr = dupSubUtil(root.right,subtrees);
if (rStr==(s))
return s;

// Serialize current subtree
s = s + root.data + lStr + rStr;

// If current subtree already exists in hash
// table. [Note that size of a serialized tree
// with single node is 3 as it has two marker
// nodes.
if (s.length > 3 && subtrees.has(s))
return "";

return s;
}

//Function to find if the Binary Tree contains duplicate
//subtrees of size 2 or more
function dupSub(root)
{
let subtrees=new Set();

return dupSubUtil(root,subtrees);
}

let root = new Node('A');
root.left = new Node('B');
root.right = new Node('C');
root.left.left = new Node('D');
root.left.right = new Node('E');
root.right.right = new Node('B');
root.right.right.right = new Node('E');
root.right.right.left= new Node('D');
let str = dupSub(root);
if(str==(""))
document.write(" Yes ");
else
document.write(" No ");

// This code is contributed by unknown2108
</script>

Output:

Yes

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